20160920, 15:08  #1 
Jul 2014
2^{2}·3·37 Posts 
2 holes in bizarre theorem about composite Mersenne Numbers
Hi again,
I've worked out a bizarre theorem about when Mersenne Numbers are composite. There's a couple of holes left in it which I don't think mean it's wrong. It's just that I've worked so hard on the problem, I can't see the wood for the trees at the moment  ie what's obvious and what isn't, I'm wondering if someone can help with the holes because I think someone can : Hole 1 : If q = 2kp + 1 divides 2^p  1 then q cannot divide any number of the form 2^c  1 where c is composite and less than p. Hole 2 : If 2^k = sq + r where q is factor of the Mersenne Number 2^p  1 then r cannot be equal to p. Other than these two things I've got a theorem which I'd be happy to share when I've written it up properly. 
20160920, 15:28  #2  
"Forget I exist"
Jul 2009
Dumbassville
2^{6}·131 Posts 
Quote:
Last fiddled with by science_man_88 on 20160920 at 15:35 

20160920, 15:38  #3 
Jul 2014
2^{2}·3·37 Posts 
Massive thanks. Will only post again if I can't assimilate what you've expressed into my work.

20160920, 15:42  #4 
Jul 2014
2^{2}·3·37 Posts 
Except for this one in which I want to make clear that I'll post the theorem soon.

20160920, 15:46  #5 
"Forget I exist"
Jul 2009
Dumbassville
8384_{10} Posts 

20160920, 16:29  #6 
Jul 2014
2^{2}×3×37 Posts 
I can't wait to get it written up properly.
I'm not sure if it's all that amazing or more importantly whether it's useful. If someone could type it up in latex for me, that would be fab because then it would be easier to read and I could be sure that it has been understood properly. The bizarre theorem ; This is I suspect part of a much stronger result. This is for when q = 2p + 1 q = 2p + 1 divides 2^p  1 if and only if (2^p) * (cos(pi/q)) * capital_pi( (k = 1 to p  1), cos((2kpi)/q) = 1 Would be most interested to know what people think of it. 
20160920, 17:24  #7  
"Forget I exist"
Jul 2009
Dumbassville
20C0_{16} Posts 
Quote:


20160920, 17:28  #8 
Aug 2006
13347_{8} Posts 
This is what you seem to mean:
Theorem. Let p be a prime number and let q = 2p + 1. Then q divides 2^p  1 if and only if \[ 2^p \cdot \cos(\pi/q) \cdot \prod_{k = 1}^{p  1} \cos((2k\pi)/q) = 1 \] But it doesn't seem to be right. p = 79 and p = 83 both give 1, but only one of those has the required divisibility property. p = 5 gives 1 but doesn't have the property. Perhaps I've misinterpreted something though, or maybe my calculations have gone awry (though I checked my results with two programs). Computationally it's worthless  products of trig functions over huge ranges are much, much slower than naive modular exponentiation. There may be some aesthetic value to such a theorem, though. But you'll need to work the bugs out. Last fiddled with by CRGreathouse on 20160920 at 17:29 
20160920, 18:02  #9  
"Forget I exist"
Jul 2009
Dumbassville
10000011000000_{2} Posts 
Quote:
http://www.purplemath.com/modules/idents.htm states: cos(x)cos(y) = 1/2[cos(xy)+cos(x+y)] could this narrow it down in complexity since p1 is even ( at least assuming p is an odd prime) they could all be paired up inside the product to form a sum. admittedly we know which p can have 2p+1 divide them but it might be nice to explore a different angle of it. Last fiddled with by science_man_88 on 20160920 at 18:06 

20160920, 18:14  #10 
Jul 2014
2^{2}×3×37 Posts 
I was totally sure it was right after the holes were revealed to be trivial.
Now that it's been doubted I'm fairly sure it's true. I've got a full proof. I'm open to the idea I made a sign error or took only a positive square root when I should have taken both. If it's true for 5 that's because like I said I think it's part of a stronger result. 
20160920, 18:15  #11 
Jul 2014
2^{2}·3·37 Posts 
Oh yeah, and thanks for typing out the latex for it.

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